#title: Integral de superfície 20 A

#author: Smirnov, Irene

#usepackage: \usepackage{tikz}

#let:

f def_pi = pi :: %pi;
n a = [2,3];
n b = [-2,-1,2];
n m = [-2,-1,2];
n r = [2,3];
n aa = 4*#a*#a*#m*#m;
n m2 = 2*#m-1;
n m3 = 2*#m-2;
n mm = 4*#m-2;
n mmm = 4*#m-4;
n rr = #r*#r;
n m5 = 4*(4*#m -2)*#a*#a*#m*#m;

f r3 = 1+4*#a*#a*#m*#m*#r^(4*#m-2);
f I = ratsimp(((#r3)^(#b+3/2)-1)/(2*(4*#m-2)*#a*#a*#m*#m*(#b+3/2)));

#question:

\noindent Calcule o integral de superfície
$$
I=\int\int_{z=#a(x^2+y^2)^{#m}, \; x^2+y^2\leq #rr} (1+#aa (x^2+y^2)^{#m2})^{#b} (x^2+y^2)^{#m3}dS.
$$

#sugestion:

#resolution:  Pela definição do integral de superfície temos
$$
 I=\int\int_{x^2+y^2\leq #rr} (1+#aa (x^2+y^2)^{#m2})^{#b} (x^2+y^2)^{#m3}
 \sqrt{1+((#a (x^2+y^2)^{#m})'_x)^2+ ((#a (x^2+y^2)^{#m})'_y)^2}dxdy
$$
$$
 =\int\int_{x^2+y^2\leq #rr} (1+#aa (x^2+y^2)^{#m2})^{#b} (x^2+y^2)^{#m3} \sqrt{1+#aa (x^2+y^2)^{#m2}}dxdy.
$$
A seguinte figura representa o domínio de integração no plano $xy$.
\begin{center}
\begin{tikzpicture}[scale=0.3]
\fill[gray!30] (0,0) circle (#r);
\draw[black] (0,0) circle (#r);
\draw[->](-7,0)--(7,0) node[right]{$x$};
\draw[->](0,-7)--(0,7) node[above]{$y$};
\foreach \x/\xtext in { 2/2,  4/4,  6/6, }
\draw[shift={(\x,0)}] (0pt,0.3pt) -- (0pt,-0.3pt) node[below] {$\xtext$};
\foreach \x/\xtext in {-6/6, -4/4, -2/2 }
\draw[shift={(\x,0)}] (0pt,0.3pt) -- (0pt,-0.3pt) node[below] {$-\xtext$};
\foreach \y/\ytext in { 2/2, 4/4, 6/6}
\draw[shift={(0,\y)}] (0.3pt,0pt) -- (-0.3pt,0pt) node[left] {$\ytext$};
\foreach \y/\ytext in {-6/6, -4/4, -2/2}
\draw[shift={(0,\y)}] (0.3pt,0pt) -- (-0.3pt,0pt) node[left] {$-\ytext$};
\foreach \x in {-7,...,7}
\draw (\x ,-1pt) -- (\x ,1pt);
\foreach \y in {-7,...,7}
\draw (-1pt,\y ) -- (1pt,\y );
\end{tikzpicture}
\end{center}

Introduzindo as coordenadas polares $x=r\cos\phi$ e $y=r\sin\phi$, obtemos
$$
 I=\int_0^{2\pi}d\phi \int_0^{#r} (1+#aa r^{#mm})^{#b}r^{#mmm}\sqrt{1+#aa r^{#mm}}rdr.
$$
Fazendo $u=1+#aa r^{#mm}$, obtemos
$$
 I=\frac{2\pi}{#m5} \int_1^{#r3} u^{#b+\frac{1}{2}}du = #I \pi .
$$


#result:


#Verification: